For one mole of a Van der Waals' gas when b = 0 and T = 300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a (atm L $m o l^{- 2}$ )

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4.5

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1.5

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Solution

The correct option is C 1.5
Van der Waals' equation of state is given as:
$(p + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T$

Given,
n = 1 and b = 0
$(p + \frac{a}{V^{2}}) V = R T$
$\Rightarrow p V = R T - \frac{a}{V}$ ....(i)

This is an equation of straight line between pV and $\frac{1}{V}$ whose slope - a.

Slope of the given graph is,
$- a = \frac{20.1 - 21.6}{3 - 2} = - 1.5$
$\Rightarrow$ a = 1.5

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For one mole of a Van der Waals' gas when b = 0 and T = 300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a (atm L mol−2)

The correct option is C 1.5Van der Waals' equation of state is given as: (p+n2aV2)(V−nb)=nRT Given, n = 1 and b = 0 (p+aV2)V=RT ⇒pV=RT−aV ....(i) This is an equation of straight line between pV and1V whose slope - a. Slope of the given graph is, −a=20.1−21.63−2=−1.5 ⇒ a = 1.5

For one mole of a Van der Waals' gas when b = 0 and T = 300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a (atm L $m o l^{- 2}$ )